Convergence of generic p-series












0












$begingroup$


Consider the generic p-series:



$sum_{n=1}^{infty}frac{1}{n^{p}}$



Let the increment function be: $n= n+K$



where $K$ is some integer constant. For $K=1$ its simple p-series and that diverges for any $p<1$.



Can we have some integer value of $K$ beyond which the series converges for $p<1$. Specifically for $ 1/2 < p <1.$






real-analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $displaystylesum_{n=1}^{infty}frac{1}{(n+K)^p}$ converges iff $p>1$ (regardless of $K$). Did you mean something else?
    $endgroup$
    – metamorphy
    Dec 18 '18 at 10:57












  • $begingroup$
    Ah it didn't sorry. The question is what if we consider every $K^{th}$ term for the values of $n$. Does the same constraint still hold true? or can we for some $K$ have the series converge even if $p<1$. I would thus rewrite the question as $displaystylesum_{n=1}^{infty}frac{1}{(n*K)^p}$
    $endgroup$
    – TheoryQuest1
    Dec 18 '18 at 11:13








  • 1




    $begingroup$
    Then it is $displaystylefrac{1}{K^p}sum_{n=1}^{infty}frac{1}{n^p}$. With the same outcome.
    $endgroup$
    – metamorphy
    Dec 18 '18 at 11:16
















0












$begingroup$


Consider the generic p-series:



$sum_{n=1}^{infty}frac{1}{n^{p}}$



Let the increment function be: $n= n+K$



where $K$ is some integer constant. For $K=1$ its simple p-series and that diverges for any $p<1$.



Can we have some integer value of $K$ beyond which the series converges for $p<1$. Specifically for $ 1/2 < p <1.$






real-analysis







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $displaystylesum_{n=1}^{infty}frac{1}{(n+K)^p}$ converges iff $p>1$ (regardless of $K$). Did you mean something else?
    $endgroup$
    – metamorphy
    Dec 18 '18 at 10:57












  • $begingroup$
    Ah it didn't sorry. The question is what if we consider every $K^{th}$ term for the values of $n$. Does the same constraint still hold true? or can we for some $K$ have the series converge even if $p<1$. I would thus rewrite the question as $displaystylesum_{n=1}^{infty}frac{1}{(n*K)^p}$
    $endgroup$
    – TheoryQuest1
    Dec 18 '18 at 11:13








  • 1




    $begingroup$
    Then it is $displaystylefrac{1}{K^p}sum_{n=1}^{infty}frac{1}{n^p}$. With the same outcome.
    $endgroup$
    – metamorphy
    Dec 18 '18 at 11:16














0












0








0





$begingroup$


Consider the generic p-series:



$sum_{n=1}^{infty}frac{1}{n^{p}}$



Let the increment function be: $n= n+K$



where $K$ is some integer constant. For $K=1$ its simple p-series and that diverges for any $p<1$.



Can we have some integer value of $K$ beyond which the series converges for $p<1$. Specifically for $ 1/2 < p <1.$






real-analysis







share|cite|improve this question









$endgroup$




Consider the generic p-series:



$sum_{n=1}^{infty}frac{1}{n^{p}}$



Let the increment function be: $n= n+K$



where $K$ is some integer constant. For $K=1$ its simple p-series and that diverges for any $p<1$.



Can we have some integer value of $K$ beyond which the series converges for $p<1$. Specifically for $ 1/2 < p <1.$






real-analysis




real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 18 '18 at 10:27









TheoryQuest1TheoryQuest1

1417




1417








  • 1




    $begingroup$
    $displaystylesum_{n=1}^{infty}frac{1}{(n+K)^p}$ converges iff $p>1$ (regardless of $K$). Did you mean something else?
    $endgroup$
    – metamorphy
    Dec 18 '18 at 10:57












  • $begingroup$
    Ah it didn't sorry. The question is what if we consider every $K^{th}$ term for the values of $n$. Does the same constraint still hold true? or can we for some $K$ have the series converge even if $p<1$. I would thus rewrite the question as $displaystylesum_{n=1}^{infty}frac{1}{(n*K)^p}$
    $endgroup$
    – TheoryQuest1
    Dec 18 '18 at 11:13








  • 1




    $begingroup$
    Then it is $displaystylefrac{1}{K^p}sum_{n=1}^{infty}frac{1}{n^p}$. With the same outcome.
    $endgroup$
    – metamorphy
    Dec 18 '18 at 11:16














  • 1




    $begingroup$
    $displaystylesum_{n=1}^{infty}frac{1}{(n+K)^p}$ converges iff $p>1$ (regardless of $K$). Did you mean something else?
    $endgroup$
    – metamorphy
    Dec 18 '18 at 10:57












  • $begingroup$
    Ah it didn't sorry. The question is what if we consider every $K^{th}$ term for the values of $n$. Does the same constraint still hold true? or can we for some $K$ have the series converge even if $p<1$. I would thus rewrite the question as $displaystylesum_{n=1}^{infty}frac{1}{(n*K)^p}$
    $endgroup$
    – TheoryQuest1
    Dec 18 '18 at 11:13








  • 1




    $begingroup$
    Then it is $displaystylefrac{1}{K^p}sum_{n=1}^{infty}frac{1}{n^p}$. With the same outcome.
    $endgroup$
    – metamorphy
    Dec 18 '18 at 11:16








1




1




$begingroup$
$displaystylesum_{n=1}^{infty}frac{1}{(n+K)^p}$ converges iff $p>1$ (regardless of $K$). Did you mean something else?
$endgroup$
– metamorphy
Dec 18 '18 at 10:57






$begingroup$
$displaystylesum_{n=1}^{infty}frac{1}{(n+K)^p}$ converges iff $p>1$ (regardless of $K$). Did you mean something else?
$endgroup$
– metamorphy
Dec 18 '18 at 10:57














$begingroup$
Ah it didn't sorry. The question is what if we consider every $K^{th}$ term for the values of $n$. Does the same constraint still hold true? or can we for some $K$ have the series converge even if $p<1$. I would thus rewrite the question as $displaystylesum_{n=1}^{infty}frac{1}{(n*K)^p}$
$endgroup$
– TheoryQuest1
Dec 18 '18 at 11:13






$begingroup$
Ah it didn't sorry. The question is what if we consider every $K^{th}$ term for the values of $n$. Does the same constraint still hold true? or can we for some $K$ have the series converge even if $p<1$. I would thus rewrite the question as $displaystylesum_{n=1}^{infty}frac{1}{(n*K)^p}$
$endgroup$
– TheoryQuest1
Dec 18 '18 at 11:13






1




1




$begingroup$
Then it is $displaystylefrac{1}{K^p}sum_{n=1}^{infty}frac{1}{n^p}$. With the same outcome.
$endgroup$
– metamorphy
Dec 18 '18 at 11:16




$begingroup$
Then it is $displaystylefrac{1}{K^p}sum_{n=1}^{infty}frac{1}{n^p}$. With the same outcome.
$endgroup$
– metamorphy
Dec 18 '18 at 11:16










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